Problem: A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has a lower base radius of 6 inches, an upper base radius of 3 inches, and a height of 4 inches, what is its lateral surface area?  (The lateral surface area of a cone or frustum is the curved surface excluding the base(s).)

[asy]size(200);
import three; defaultpen(linewidth(.8)); currentprojection = orthographic(0,-3,0.5); pen dots = linetype("0 3") + linewidth(1);
real h = 2.3, ratio = (91-24)/(171-24);
picture p1, p2; /* p1 is left-hand picture */
triple A = (0,0,0), B = (0,0,h); draw(p1,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p1,(-1,0,0)..(0,1,0)..(1,0,0),dots); draw(p1,(-1,0,0)--B--(1,0,0));
add(p1);

triple vlift = (0,0,0.5);

path3 toparc1 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((-1,0,0)..(0,1,0)..(1,0,0)), toparc2 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((1,0,0)..(0,-1,0)..(-1,0,0));
draw(p2,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p2,(-1,0,0)..(0,1,0)..(1,0,0),dots);

draw(p2,(-1,0,0)--ratio*(-1,0,0)+(1-ratio)*B^^ratio*(1,0,0)+(1-ratio)*B--(1,0,0));

draw(p2,shift(vlift)*(ratio*(-1,0,0)+(1-ratio)*B--B--ratio*(1,0,0)+(1-ratio)*B));

draw(p2,toparc1--toparc2); draw(p2,shift(vlift)*toparc1,dots); draw(p2,shift(vlift)*toparc2);

draw(p2,shift(vlift)*((1-ratio)*B--B),linewidth(0.7)); dot(p2,shift(vlift)*((1-ratio)*B),linewidth(1.5));
label(p2,"frustum",(0,0,h/4));

add(shift((3.4,0,0))*p2);

[/asy]
Solution: We start by drawing the frustum.  Let the top and bottom circles have centers $O_1$ and $O_2$ respectively, and label points $A$ and $B$ on the circumferences as shown such that $O_1$, $O_2$, $A$, and $B$ lie in the same plane.

[asy]
unitsize(0.5 cm);

import geometry; defaultpen(linewidth(.8)+fontsize(10));
label("$O_1$",(0,4),W); label("$O_2$",(0,0),SW); label("$B$",(6,0),SE); label("$A$",(3,4),NE);
draw((3,4)--(0,4)--(0,0)--(6,0));
draw(scale(1,.2)*arc((0,0),6,0,180),linetype("2 4"));
draw(scale(1,.2)*arc((0,0),6,180,360));
draw(scale(1,.2)*arc((0,20),3,0,180));
draw(scale(1,.2)*arc((0,20),3,180,360));
draw((6,0)--(3,4)); draw((-6,0)--(-3,4));
label("6",(3,0),S); label("4",(0,2),W); label("3",(1.5,4),N);

[/asy]

Because the frustum was cut from a right circular cone, $\angle AO_1O_2$ and $\angle BO_2O_1$ are both right angles.  We drop a perpendicular from $A$ to $\overline{O_2B}$ and let the intersection point be $X$.  Then $O_1AXO_2$ is a rectangle and \[XB=O_2B-O_1A=6-3=3.\]Pythagorean theorem on right $\triangle AXB$ gives \[AB=\sqrt{AX^2 + BX^2}=\sqrt{4^2+3^2}=5.\]Thus the slant height of the frustum is 5.

Extend $\overline{O_1O_2}$ and $\overline{AB}$ above the frustum, and let them intersect at point $C$. $C$ is the tip of the full cone that the frustum was cut from.  To compute the lateral surface area of the frustum, we compute the lateral surface area of the full cone and subtract off the lateral surface area of the smaller cone that was removed.

[asy]
unitsize(0.5 cm);

import geometry; defaultpen(linewidth(.8)+fontsize(10));
label("$O_1$",(0,4),W); label("$O_2$",(0,0),SW); label("$B$",(6,0),SE); label("$A$",(3,4),NE);
draw((3,4)--(0,4)--(0,0)--(6,0)); draw((3,4)--(0,8)--(-3,4)); draw((0,4)--(0,8)); label("$C$",(0,8),NE);
draw(scale(1,.2)*arc((0,0),6,0,180),linetype("2 4"));
draw(scale(1,.2)*arc((0,0),6,180,360));
draw(scale(1,.2)*arc((0,20),3,0,180),linetype("2 4"));
draw(scale(1,.2)*arc((0,20),3,180,360));
draw((6,0)--(3,4)); draw((-6,0)--(-3,4));
label("6",(3,0),S); label("4",(0,2),W); label("3",(1.5,4),N); label("5",(4.5,2),NE); [/asy]

To find the height of the whole cone, we take a vertical cross-section of the cone that includes $O_1$, $O_2$, $A$, and $B$.  This cross-section is an isosceles triangle.

[asy]
unitsize(0.5 cm);

defaultpen(linewidth(.8)+fontsize(10));
draw((0,0)--(12,0)--(6,8)--cycle); draw((6,0)--(6,8)); draw((6,4)--(9,4));
label("$B$",(12,0),E); label("$C$",(6,8),NE); label("$O_1$",(6,4),W); label("$O_2$",(6,0),SW); label("$A$",(9,4),E);

label("6",(9,0),S); label("3",(7.5,4),S); label("4",(6,2),W); label("5",(10.5,2),NE);

[/asy]

$\triangle CO_1A$ and $\triangle CO_2B$ are similar, so \[\frac{CO_1}{CO_2} = \frac{CA}{CB}=\frac{O_1A}{O_2B}=\frac{3}{6}.\]Thus $CO_1=4$ and $CA=5$ (and we see the small removed cone has half the height of the full cone).  Also, $CB=10$.

Now we unroll the lateral surface area of the full cone.  (The desired frustum lateral area is shown in blue.)

[asy]
unitsize(0.2 cm);

import graph;
defaultpen(linewidth(.8)+fontsize(10));
fill(Arc((0,0),10,0,240)--cycle,heavycyan); fill(Arc((0,0),5,0,240)--cycle,white); fill((5,0)--(10,0)--(-5,-5*sqrt(3))--(-2.5,-2.5*sqrt(3))--cycle,white);
draw(Arc((0,0),10,0,240)); draw(Arc((0,0),5,0,240));
draw(Arc((0,0),10,240,360),linetype("2 4")); draw(Arc((0,0),5,240,360),linetype("2 4"));

draw((10,0)--(0,0)--(-5,-5*sqrt(3)));

label("$C$",(0,0),SE); label("$A$",(5,0),SE); label("$B$",(10,0),SE); label("10",(-2.5,-2.5*sqrt(3)),SE);

[/asy]

When unrolled, the full cone's lateral surface area is a sector whose arc length is the cone's base perimeter and whose radius is the cone's slant height.  So, the sector has arc length $2\cdot \pi \cdot 6 = 12\pi$ and radius $10$.  A full circle with radius 10 has arc length $2\cdot \pi \cdot 10 = 20\pi$, so the sector has $\frac{12\pi}{20\pi}=\frac{3}{5}$ of the circle's arc length and thus has 3/5 of the circle's area.  Thus, the full cone has lateral surface area \[\frac{3}{5}\pi (10^2) = 60\pi.\]Similarly, the small removed cone's lateral surface area is a sector with radius 5 and arc length $2\cdot \pi \cdot 3 = 6\pi$ (which is $3/5$ of the arc length of a full circle with radius 5), so its lateral surface area is \[\frac{3}{5}\pi (5^2)=15\pi.\]The lateral surface area of the frustum, in blue, is the full cone's lateral surface area minus the small removed cone's lateral surface area, which is \[60\pi-15\pi=\boxed{45\pi}.\]